On the optimal control of linear complementarity systems Vieira - - - PowerPoint PPT Presentation

IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

On the optimal control of linear complementarity systems

Alexandre Vieira1 - Bernard Brogliato1 - Christophe Prieur2

1 Univ. Grenoble Alpes, INRIA Grenoble - 2 Univ. Grenoble Alpes, GIPSA-Lab

22nd May 2018

alexandre.vieira@inria.fr, bernard.brogliato@inria.fr, christophe.prieur@gipsa-lab.fr.

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Introduction

min c(xu, u) s.t. ˙ xu(t) = f (xu(t), u(t)), (xu(t), u(t)) ∈ S(t), xu(0) = x0, xu(T) free.

x(t) c(u2) c(u3) c(u4) c(u5) c(u1) T 2 / 38

IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Introduction

Problem: C(u) = T (x(t)⊺Qx(t) + u(t)⊺Uu(t) + v(t)⊺Vv(t)) dt → min such that:      ˙ x(t) = Ax(t) + Bv(t) + Fu(t) 0 ≤ v(t) ⊥ w(t) = Cx(t) + Dv(t) + Eu(t) ≥ 0 x(0) = x0, x(T) free where T > 0, x : [0, T] → Rn absolutely continuous, v : [0, T] → Rm, u : [0, T] → Rmu, A, B, C, D, E, F, Q, V and U matrices of according dimensions, U supposed symmetric positive definite, Q and V positive semi-definite. Motivation: Mechanics, Electronic Circuits, Chemical reactions

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

A difficult problem

0 ≤ v(t) ⊥ Cx(t) + Dv(t) + Eu(t) ≥ 0 Existence of optimal solution not proved (classical Fillipov theory does not apply here due to lack of convexity). Cesari (2012), Theorem 9.2i and onwards Special cases arise when E = 0 and D P-matrix : switching modes are activated when the state reaches some threshold defined by the complementarity conditions. Georgescu et al. (2012), Passenberg et al. (2013) Since u is also involved = ⇒ mixed constraints; makes use of non-smooth

• analysis. Clarke and De Pinho (2010)

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Direct Method

A first way to compute numerically an approximate solution: direct method. min

N

• i=0

xiQxi + uiUui s.t.      xi+1 = xi + h(Axi + Bvi + Eui) 0 ≤ vi ⊥ Cxi + Dvi + Eui ≥ 0 x0 fixed = ⇒ Mathematical Program with Equilibrium Constraints (MPEC). Since it is a convex cost with linear constraints: there exist a solution for any fixed h.

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Direct Method

2 ways to solve numerically this problem:

1 Cost penalization on the complementarity constraint v⊺(Cx + Dv + Eu) = 0:

didn’t work well on examples.

2 Relaxation of the complementarity: choose a sequence εk ≥ 0 converging to 0

and create a sequence of optimization problems where the constraint v⊺(Cx + Dv + Eu) = 0 is replaced by v⊺(Cx + Dv + Eu) ≤ εk: works well.

[1] S. Leyffer, G. López-Calva, and J. Nocedal. Interior methods for mathematical programs with complementarity constraints. SIAM Journal on Optimization, 17(1):52–77, 2006. [2] C. Kanzow and A. Schwartz. A new regularization method for mathematical programs with complementarity constraints with strong convergence properties. SIAM Journal on Optimization, 23(2):770–798, 2013.

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Direct Method

Denote x1(t) = t

0 i1(s)ds + x1(0) (the charge of

the capacitor, in coulomb) and x2(t) = i2(t) (the electric current, in ampere). Then the evolution of this system is described as:                      ˙ x1(t) = −1 RC x1(t) + x2(t) − 1 R λ(t) + 1 R u2(t), ˙ x2(t) = −1 LC x1(t) − 1 Lλ(t) + 1 L(u2(t) − u1(t)), 0 ≤ λ(t) ⊥ 1 RC x1(t) − x2(t) + 1 R λ(t) − 1 R u2(t) ≥ 0, The constants are chosen as R = 10Ω, C = 80 000µF, and L = 2H. L i2 D λ C i1 R u1 u2

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Direct Method

min 1 [Rx2(t)2 + u1(t)2 + u2(t)2]dt s.t.                                ˙ x1(t) = −1 RC x1(t) + x2(t) − 1 R λ(t) + 1 R u2(t), ˙ x2(t) = −1 LC x1(t) − 1 Lλ(t) + 1 L(u2(t) − u1(t)), 0 ≤ λ(t) ⊥ 1 RC x1(t) − x2(t) + 1 R λ(t) − 1 R u2(t) ≥ 0, x(0) = (200C, 50A) x(1) = (50C, 0A) L x2 D λ C ˙ x1 R u1 u2

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Direct Method

0.0 0.2 0.4 0.6 0.8 1.0 t −150 −100 −50 50 100 150 200

x1 x2

λ always naught!

0.0 0.2 0.4 0.6 0.8 1.0 t −600 −400 −200 200 400 600 800 1000

u1 u2

L x2 D λ C ˙ x1 R u1 u2

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Direct Method

min 10

• x(t)2

2 + u(t)2

dt, s.t.                            ˙ x(t) =   1 1   x(t) +   v(t) + u(t)   , 0 ≤ v(t) ⊥

• 1
• x(t) + u(t) ≥ 0,

a.e. on [0, 10] x(0) =   −2 1 −1   , x(T) free, Resolution with Direct Method and relaxation of the complementarity. Library used: IPOPT and

• CasADI. h = 10−3.

2 4 6 8 10 t −4 −3 −2 −1 1 2 3 4 5

x1 x2 x3

2 4 6 8 10 t 200 400 600 800 1000 1200 1400 1600 1800

v1 10 / 38

IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Why do we bother ?

Suppose an optimal solution exists = ⇒ Search for necessary conditions. Two reasons for that: Useful for analyzing the solution (continuity, sensitivity...) The direct method mostly works fine! But very slow for high precision or big

• systems. Possible pseudominima?

Really general necessary conditions were obtained in [1]. But as such, they are not really practical (complicated hypothesis, really general equations...). Can it be enhanced in the case of LCS?

[1] L. Guo and J. J. Ye. Necessary optimality conditions for optimal control problems with equilibrium constraints (2016).

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Weak stationarity

Define S = {(x, u, v)|0 ≤ v ⊥ Cx + Dv + Eu ≥ 0} and the partition of {1, ..., m}: I 0+

t

(x, u, v) = {i | vi(t) = 0 < (Cx(t) + Dv(t) + Eu(t))i} I +0

t

(x, u, v) = {i | vi(t) > 0 = (Cx(t) + Dv(t) + Eu(t))i} I 00

t (x, u, v) = {i | vi(t) = 0 = (Cx(t) + Dv(t) + Eu(t))i}

Theorem Let (x∗, u∗, v∗) be a local minimizer of radius R(·). Suppose Im(C) ⊆ Im(E). Then there exist an absolutely continuous function p : [0, T] → Rn and measurable functions λG : R → Rm, λH : R → Rm such that the following conditions hold:

1 the transversality condition: p(T) = 0

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Weak stationarity

Theorem

2 the Weierstrass condition for radius R: for almost every t ∈ [t0, t1],

(x∗(t), u, v) ∈ S,

• u

v

• −

u∗(t) v∗(t)

• < R(t)

= ⇒ p(t), Ax∗(t) + Bv + Fu) − 1 2 (x∗(t)⊺Qx∗(t) + u⊺Uu + v⊺Vv) ≤ p(t), Ax∗(t) + Bv∗(t) + Fu∗(t)) − 1 2 (x∗(t)⊺Qx∗(t) + u∗(t)⊺Uu∗(t) + v∗(t)⊺Vv∗(t))

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Weak stationarity

Theorem

3 the Euler adjoint equation: for almost every t ∈ [0, T],

˙ p(t) = −A⊺p(t) + Qx∗(t) − C ⊺λH(t) 0 = F ⊺p(t) − Uu∗(t) + E ⊺λH(t) 0 = B⊺p(t) + λG(t) + D⊺λH(t) 0 = λG

i (t), ∀i ∈ I +0 t

(x∗(t), u∗(t), v∗(t)) 0 = λH

i (t), ∀i ∈ I 0+ t

(x∗(t), u∗(t), v∗(t)) Remark : One can also prove that these conditions are, in one sense, sufficient.

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Euler equation

How can one solve the following BVP? ˙ x = Ax + Bv + Fu ˙ p = −A⊺p + Qx − C ⊺λH 0 = F ⊺p − Uu + E ⊺λH 0 = B⊺p + λG + D⊺λH 0 = λG

i (t), ∀i ∈ I +0 t

(x(t), u(t), v(t)) 0 = λH

i (t), ∀i ∈ I 0+ t

(x(t), u(t), v(t)) x0 = x(0), 0 = p(T)

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Euler equation

How can we solve the following BVP? ˙ x = Ax + Bv + Fu ˙ p = −A⊺p + Qx − C ⊺λH 0 = F ⊺p − Uu + E ⊺λH → isolate u 0 = B⊺p + λG + D⊺λH → isolate λG 0 = λG

i (t), ∀i ∈ I +0 t

(x(t), u(t), v(t)) 0 = λH

i (t), ∀i ∈ I 0+ t

(x(t), u(t), v(t)) x0 = x(0), 0 = p(T)

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Strong stationarity

0 = λG

i (t), ∀i ∈ I +0 t

(x(t), u(t), v(t)) 0 = λH

i (t), ∀i ∈ I 0+ t

(x(t), u(t), v(t)) We miss a piece of information: what happens on I 00

t

? Proposition Let (x∗, u∗, v∗) be a local minimizer and suppose E invertible. Then (x∗, u∗, v∗) is strongly stationary, meaning: λG

i (t) ≥ 0, λH i (t) ≥ 0, ∀i ∈ I 00 t (x(t), u(t), v(t))

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Strong stationarity

0 = λG

i (t),

∀i ∈ I +0

t

(x(t), u(t), v(t)) 0 = λH

i (t),

∀i ∈ I 0+

t

(x(t), u(t), v(t)) λG

i (t) ≥ 0, λH i (t) ≥ 0,

∀i ∈ I 00

t (x(t), u(t), v(t))

Almost like a linear complementarity problem!

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Strong stationarity

Theorem Let (x∗, u∗, v∗) be a local minimizer and suppose E invertible. Fix an arbitrary r > 0. Then there exist an arc p and measurable functions β : [0, T] → Rm, ζ : [0, T] → R such that, u∗(t) = U−1 (F ⊺p(t) + E ⊺β(t) − (ζ(t) + r)E ⊺v∗(t)) and: ˙ x ˙ p

• = A

x p

• + B

β v∗

• 

        0 ≤ β v∗

• ⊥ D

β v∗

• + C

x p

• ≥ 0

β ≥ rv∗ x(0) = x0, p(T) = 0

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

How to solve a BVP LCS

˙ x ˙ p

• = A

x p

• + B

β v

• 

          0 ≤ β v

• ⊥ D

β v

• + C

x p

• ≥ 0

β ≥ rv x(0) = x0, p(T) = 0 Numerically, we usually do shooting: find the good p(0) = p0 such that the computed solution p(t; p0) complies with p(T; p0) = 0: nonsmooth Newton method. Need for an initial guess close enough How to compute a sensitivity matrix for p(T; ·) ?

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

How to solve a BVP LCS

˙ z = Az + BΛ 0 ≤ Λ ⊥ DΛ + Cz ≥ 0 Denote Th(z) a linear Newton Approximation to the solution Λ of the LCP. Then, a linear Newton approximation for the solution map z(T, ·) can be obtained by solving the DI in matrix function: ˙ J(t) ∈ AJ(t) + (co Th(z(t; ξ)))J(t), J(0) = I ... But it supposes that B SOL(D, Cz) is a singleton for all z ∈ R2n (which we can not prove).

JS Pang, D. Stewart, Solution dependence on initial conditions in differential variational inequalities (2009)

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

The indirect method

The indirect method consists in solving numerically the necessary conditions (also sufficient in this case). Since it is based on a shooting method (involving a nonsmooth Newton method), the equations are solved in two steps:

1 One solves, roughly, the optimal control problem with the Direct method, in

• rder to get a rough idea of the solution.

2 One refines the solution by solving the necessary conditions, giving the solution

• f the Direct method as an initial guess.

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

A 1D example

˙ x = ax + bv + fu 0 ≤ v ⊥ dv + eu ≥ 0 x(0) = x0 We can show that the (strong) stationary solution in this case is given by: p(t) =

• cosh(√γt) − a

√γ sinh(√γt)

• p(0) + sinh(√γt)

√γ x(0) p(0) = − sinh(√γT) √γ cosh(√γT) − a sinh(√γT)x(0). u(t) =

• fp(t)

if efp(0) ≥ 0,

• f − eb

d

• p(t)

if efp(0) ≤ 0. x(t) = ˙ p(t) + ap(t).

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

A 1D example

2 4 6 8 10 t −1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2

• Analy. x
• Num. x

2 4 6 8 10 t −0.5 0.0 0.5 1.0 1.5 2.0

• Analy. u
• Num. u

Figure: Solution via indirect method : state x and control u, on [0, 10]. a = 1, b = 0.5, d = 1, e = −2, f = 3, x(0) = −1. Initial guess with direct method and 300

• nodes. Indirect method with 10 000 nodes and 20 intervals of shooting. Obtained in 54s.

(In order to have this same precision with the direct method : 453s.)

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Compare direct and indirect method

Let us compare the time of computation using the Direct Method and the Hybrid Approach (rough direct + refinements with indirect) in this example: min 1

• x(t)2

2 + 25u(t)2 2

• dt,

s.t.                  ˙ x(t) = 1 2 2 1

• x(t) +

−1 1 −1 1

• v(t) +

1 3 2 1

• u(t),

0 ≤ v(t) ⊥ 3 −1 −2

• x(t) + v(t) +

1 −1 −1 2

• u(t) ≥ 0,

a.e. on [0, 1] x(0) = − 1

2

1

• , x(T) free,

(1)

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Compare direct and indirect method

hD Time spent (s) 10−2 1.31 10−3 37.50 10−4 400.65 10−5 ∞ 10−6 ∞

Table: Time spent With Direct Method

Parameters Time spent (s) hD = 10−1, hI = 10−2, nS = 5 1.39 hD = 10−1, hI = 10−3, nS = 10 11.26 hD = 10−2, hI = 10−4, nS = 20 97.56 hD = 10−3, hI = 10−5, nS = 50 1 298.62 hD = 10−4, hI = 10−6, nS = 100 32 163.36

Table: Time spent with Hybrid approach

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Minimal time

Let us review a different problem: T ∗ = min T(x, u, v) s.t.          ˙ x(t) = Ax(t) + Bv(t) + Fu(t) 0 ≤ v(t) ⊥ Cx(t) + Dv(t) + Eu(t) ≥ 0, u(t) ∈ U x(0) = x0, x(T ∗) = xf . where U is a finite union of polyhedral compact convex sets. Since u is now constrained: no more possibility to have strong stationarity and do the same manipulations. One still could have a weaker result, but not really useful as is.

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

A bang-bang property

T ∗ = min T(x, u, v) s.t.          ˙ x(t) = Ax(t) + Bv(t) + Fu(t) 0 ≤ v(t) ⊥ Dv(t) + Eu(t) ≥ 0, u(t) ∈ U x(0) = x0, x(T ∗) = xf . We just suppose that D is a P-matrix. Denote by Ω = {(u, v) ∈ U × Rm|0 ≤ v ⊥ Dv + Eu ≥ 0}, and AccΩ(x0, t) the accessible set at time t, starting from x0 with controls having values in Ω. For an index set α ⊆ {1, ..., m}, denote by Rm

α the set of points q in Rm such that

qα ≥ 0, qm\α ≤ 0, and define E −1Rm

α = {˜

u ∈ Rm|E ˜ u ∈ Rm

α } (E is not necessarily

invertible)

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

A bang-bang property

Theorem For a certain α ⊆ {1, ..., m}, denote by Eα the set: Eα =

• (u, v) ∈ Ext
• U ∩ E −1Rm

α

• × Rm| vα = 0, Dα•v + Eα•u = 0,

v ≥ 0, Dv + Eu ≥ 0

• and by E the set E =

α⊆m Eα. Then, for all t > 0 and all x0 ∈ Rn,

AccΩ(x0, t) = AccE(x0, t) Explanation when E = I : Find all extreme points of each intersection of U with each orthant of Rm. For each of these extreme points, find the solution v of the

• LCP. The optimal control can be searched as a succession of arcs taking these

values.

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

A bang-bang property

Example T ∗ = min T(x, u, v) s.t.          ˙ x(t) = ax(t) + bv(t) + fu(t), 0 ≤ v(t) ⊥ v(t) + u(t) ≥ 0, a.e. on [0, T ∗] u(t) ∈ U = [−1, 1] (x(0), x(T ∗)) = (x0, xf ), In this case, E = {(−1, 1), (0, 0), (1, 0)}. We can therefore search for the optimal solution with controls (u, v) in E. One can prove that, under complete controllability of the system, the optimal control (u∗, v∗) is constant along [0, T ∗] and equal to (−1, 1) or (1, 0).

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

A bang-bang property

This second example, close to the previous one, suggests that the hypothesis of D P-matrix can possibly be relaxed. Example T ∗ = min T(x, u, v) s.t.          ˙ x(t) = Ax(t) + Bv(t) + Fu(t), 0 ≤ v(t) ⊥ −v(t) + u(t) ≥ 0 u(t) ∈ U = [−1, 1] (x(0), x(T ∗)) = (x0, xf ), In this case, one can prove that E = {(0, 0), (1, 0), (1, 1)} also works for covering the entire accessibility set.

• u

v Ω conv

1 1 Ω

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Conclusion

First stationarity and geometrical results, that we can use analytically and numerically. Numerical algorithms working fast, even with high precision. (For those interested: the whole code is on https://gitlab.inria.fr/avieira/optLCS)

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Introduction

Why is it so hard ? Direct approach

Necessary conditions and indirect method Minimal time Conclusion

Conclusion

What is left to be done: The stationarity LCS, even in this case, still is not entirely analysed. When the dimension of the complementarity becomes high: the numerical resolution fails. Get rid of some assumptions (E invertible for the quadratic cost, D P-matrix for the minimal time...). For minimal time problem: could we algorithmically find the extreme points E? Could we extend these results with a more relaxed concept of solution (distributions)?

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Left in case

• f

Bounded Slope Condition Matrix definition

Bounded Slope Condition

There exists a positive measurable function kS such that for almost every t ∈ [0, T], the bounded slope condition holds: (x, w) ∈ Sε,R

∗

(t), (α, β) ∈ N P

S(t)(x, w) =

⇒ α ≤ kS(t)β.

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Left in case

• f

Bounded Slope Condition Matrix definition

Strong stationarity

Define, for two scalars ζ and r: A = A FU−1F ⊺ Q −A⊺

• B =

FU−1E ⊺ B − (ζ + r)FU−1E ⊺ −C ⊺ (ζ + r)C ⊺

• C =

C EU−1F ⊺ ζC ζEU−1F ⊺ − B⊺

• D =
• EU−1E ⊺

D − (ζ + r)EU−1E ⊺ ζEU−1E ⊺ − D⊺ ζD + (ζ + r)

• D⊺ − ζEU−1E ⊺
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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Left in case

• f

Bounded Slope Condition Matrix definition

Minimal time

Theorem Suppose: either C = 0,

• r D is a diagonal matrix with positive entries.

Let (x∗, u∗, v∗) be a local minimizer for the minimal time problem. Then (x∗, u∗, v∗) is W-stationary; i.e. there exist an arc p : [0, T ∗] → Rn, a scalar λ0 ∈ {0, 1} and multipliers λG, λH : [0, T ∗] → Rm such that: (λ0, p(t)) = 0 ∀t ∈ [0, T ∗] and:

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Left in case

• f

Bounded Slope Condition Matrix definition

Minimal time

Theorem ˙ p(t) = −A⊺p(t) − C ⊺λH(t) 0 = B⊺p + D⊺λH + λG 0 ∈ −F ⊺p − E ⊺λH + N C

U (u∗(t))

λG

i (t) = 0, ∀i ∈ I +0 t

(x∗, u∗, v∗) λH

i (t) = 0, ∀i ∈ I 0+ t

(x∗, u∗, v∗)

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IPL Modeliscale Alexandre Vieira - Bernard Brogliato - Christophe Prieur Left in case

• f

Bounded Slope Condition Matrix definition

Minimal time

Also, since the system is linear, we know there exist a second set of multipliers ηG and ηH such that: 0 = B⊺p + D⊺ηH + ηG 0 ∈ −F ⊺p − E ⊺ηH + N C

U (u∗(t))

ηG

i (t) = 0, ∀i ∈ I +0 t

(x∗, u∗) ηH

i (t) = 0, ∀i ∈ I 0+ t

(x∗, u∗) ηG

i ηH i = 0 or ηG i > 0, ηH i > 0, ∀i ∈ I 00 t (x∗, u∗)

... But they can be different from the corresponding λG and λH on a subset of [0, T ∗] of positive measure. λG

i

λH

i

(a) W-stationarity

ηG

i

ηH

i

(b) M-stationarity

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